partial derivatives

Don Mariano Marcos Memorial State University GRADUATE COLLEGE South La Union Campus Agoo, La Union Subject: Topic: Professor: Reporter:

Advance Calculus II Partial Derivatives, and Differentiability and the Total Differential Dr. Delia V. Eisma Eduard M. Albay

PARTIAL DERIVATIVES Differentiation of real-valued functions of n variables is reduced to the onedimensional case by treating a function of n variables as a function of one variable at a time and holding the others fixed. This leads to the concept of a partial derivative. We begin with functions with two variables.

Definition of a Partial Derivative of a Function of Two Variables

Let f be a function of two variables, x and y. the partial derivative of f with respect to x is that function, denoted by D1f, such that its function value at any point (x,y) in the domain of f is given by f ( x + ∆x, y ) − f ( x, y ) D1f(x,y) = lim ∆x →0 ∆x if this limit exists. Similarly, the partial derivative of f with respect to y is that function denoted by D2f, such that its function value at any point (x,y) in the domain of f is given by D2f(x,y) = ∆lim y →0

f ( x, y + ∆y ) − f ( x, y ) ∆y

if this limit exists. The process of computing a partial derivative is called partial differentiation. D1f is read as “D sub 1 of f,”. and this denotes the function that is the partial derivative of f with respect to the first variable. D1f(x,y) is read as “D sub 1 of f of x and y,” and this denotes the function value of D1f at the point (x,y). Other notations for D1f are f1, fx, ∂f ( x, y ) ∂ f and . Other notations for D1f(x,y) are f1(x,y), fx(x,y) and . Similarly, other ∂ x ∂x notations for D2f are f2, fy, and

∂ f ; other notations for D2f(x,y) are f2(x,y), fy(x,y) and ∂ y

∂f ( x, y ) ∂z . If z = f(x,y), we can write for D1f(x,y). A partial derivative cannot be thought ∂y ∂x of as a ratio of ∂z and ∂x because neither of these symbols has a separate meaning. You dy learned that the notation can be regarded as the quotient of two differentials when y is dx ∂z a function of the single variable x, but there is no similar interpretation for . ∂x

 Example 1. D2f(x,y) if

Apply the definition of a partial derivative to compute D1f(x,y) and f(x,y) = 3x2 – 2xy + y2.

Solution

f ( x + ∆x, y ) − f ( x, y ) ∆x 2 3( x + ∆x) − 2( x + ∆x) y + y 2 − (3 x 2 − 2 xy + y 2 ) = lim ∆x →0 ∆x 3 x 2 + 6 x∆x + 3(∆x) 2 − 2 xy − 2 y∆x + y 2 − 3 x 2 + 2 xy − y 2 ) = lim ∆x →0 ∆x

D1f(x,y) = lim

∆x →0

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6 x∆x + 3( ∆x) 2 − 2 y∆x ∆x →0 ∆x

= lim

(6 x + 3∆x − 2 y ) = ∆lim x →0 = 6x – 2y

f ( x, y + ∆y ) − f ( x, y ) ∆y

D2f(x,y) = ∆lim x →0

3 x 2 − 2 x( y + ∆y ) + ( y + ∆y ) 2 − (3 x 2 − 2 xy + y 2 ) ∆x →0 ∆y

= lim

3 x 2 − 2 xy − 2 x∆y + y 2 + 2 y∆y + (∆y ) 2 − 3 x 2 + 2 xy − y 2 ∆x →0 ∆y

= lim

− 2 x∆y + 2 y∆y − ( ∆x 2 ) ∆x →0 ∆x

= lim

( −2 x + 2 y + ∆y ) = ∆lim x →0 = -2x + 2y

If (x0,y0) is a particular point in the domain of f, then D1f(x0,y0) = lim

∆x →0

if this limit exists, and D2f(x0,y0) = ∆lim x →0

f ( x0 + ∆x, y 0 ) − f ( x0 , y 0 ) ∆x

(1)

f ( x0 , y 0 + ∆y ) − f ( x0 , y 0 ) ∆y

(2)

if this limit exists.

•

Illustration 1. We apply formula (1) to find D1f(3,-2) for the function f of example 1.

D1f(3,-2) = lim

∆x →0

f (3 + ∆x,−2) − f (3,−2) ∆x

3(3 + ∆x) 2 − 2(3 + ∆x)( −2) + (−2) 2 − (27 + 12 + 4) ∆x →0 ∆x

= lim

27 + 18 ∆x + 3(∆x) 2 + 12 + 4∆x + 4 − 43 ∆x →0 ∆x

= lim

(18 + 3∆x + 4) ∆x →0 ∆x = 22 = lim

Alternate formulas to (1) and (2) for D1f(x0,y0) and D2f(x0,y0) are given by D1f(x0,y0) = xlim →x

0

f ( x, y 0 ) − f ( x 0 , y 0 ) x − x0

(3)

f ( x0 , y ) − f ( x0 , y 0 ) y − y0

(4)

if the limit exists and D2f(x0,y0) = ylim →y

0

if this limit exists.

•

Illustration 2. We apply formula 3 to find D1f(3, -2) for the function f of Example 1

D1f(3, -2) = lim

x →3

f ( x,−2) − f (3,−2) x −3

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3 x 2 + 4 x + 4 − 43 x →3 x −3 2 3x + 4 x + 39 = lim x →3 x −3 (3 x +13 )( x − 3) = lim x →3 x −3 = lim

(3 x + 13 ) = lim x →3 = 22

•

Illustration 3. In Example 1, we showed that D1f(x,y) = 6x – 2y. Therefore, D1f(3, -2) = 18 + 4 = 22 This result agrees with those of Illustrations 1 and 2.

Comparing the definition of a partial derivative (12.3.1) with the definition of an ordinary derivative (2.1.3), we see that D1f(x,y) is the ordinary derivative of f if f is considered as a function of one variable x (i.e., y is held constant, and D2f(x,y) is the ordinary derivative of f if f is considered as a function of one variable y (and x is held constant). So the results in example 1 can be obtained more easily by applying the theorems for ordinary differentiation if y is considered constant when finding D2f(x,y).

 Example 2. Find fx(x,y) and fy(x,y) if f(x,y) = 3x3 – 4x2y + 3xy2 + sin xy2. Solution: Treating f as a function of x and holding y constant we have fx(x,y) = 9x2 – 8xy + 3y2 + y2 cos xy2 Considering f as a function of y and holding x constant we have fx(x,y) = -4x2 + 6xy + 2xy cos xy2 Geometric interpretations of the partial derivatives of a function of two variables are similar to those of a function of one variable. The graph of a function of f of two variables is a surface having equation z = f(x,y). if y is held constant (say y = y0), then z = f(x,y0) is an equation of the trace of this surface in the plane y = y0. The curve can be represented by the two equations y = y0 and z = f(x,y) (5) because the curve is the intersection of these two surfaces. Then D1f(x0,y0) is the slope of the tangent line to the curve given by (5) at the point P0(x0,y0, f(x0,y0)) in the plane y = y0. In an analogous fashion, D2f(x0,y0) represents the slope of the tangent line to the curve having equations x = x0

and

z = f(x,y)

at the point P0 in the plane x = x0.

 Example 3.

Find the slope of the tangent line to the curve of intersection of the

surface z=

1 24 − x 2 − 2 y 2 with the plane y = 2 at the point (2,2, 3 ) . 2

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Solution:

The required slope is the value of −x ∂z = 2 24 − x 2 − 2 y 2 ∂x

so at (2,2, 3 ) , −2 ∂z = 2 12 ∂x

=

∂z at the point (2,2, 3 ) . ∂x

−1 2 3

When computing a partial derivative at a particular point, it is sometimes necessary to apply formulas (1) through (4).

 Example 4. Given  xy ( x 2 − y 2 )   x2 + y2 f ( x, y ) =    0 

if (x,y) ≠ (0,0) if (x,y) = (0,0)

show that f1(0,0) = 0 and f2(0,0) = 0. Solution. We compute f1(0,0) from (3) with y0 = 0 and f2(0,0) from (4) with x0 = 0.

f ( x,0) − f (0,0) x −0 0 −0 = lim x →0 x

f1(0,0) = lim

x →0

f2(0,0) = lim y →0 = lim y →0

f (0, y ) − f (0,0) y −0

0 −0 y

0 = lim x →0

0 = lim y →0

=0

=0

 Example 5. For the function of Example 4, show:

(a)f1(0,y) = -y for all y; (b) f2(x,0) = x for all x.

Solution (a)if y ≠ 0, from (3)

f ( x, y ) − f (0, y ) x −0 2 xy ( x − y 2 ) −0 = x2 + y2 lim x →0 x

f1(0,y) = lim

x →0

= lim x →0

y( x 2 − y 2 ) x2 + y2

− y3 y2 = -y

=

(b) if x ≠ 0, from (4), f2(x,0) = lim y →0

f ( x, y ) − f ( x,0) y −0

xy ( x 2 − y 2 ) −0 = x2 + y2 lim y →0 y 2 x( x − y 2 ) = lim y →0 x2 + y2 =

x3 x2

=x

(a)because f1(0,y) = -y if y ≠ 0 and, from Example (4), f1(0,0) = 0, we can conclude that f1(0,y) = -y for all y. (b)because f2(x,0) = x if x ≠ 0 and, from Example 4, f2(0,0), then f2(x,0) = x for all x. We now extend the concept of partial derivative to functions of n variables.

Definition of the Partial Derivative of a Function of n Variables Partial Derivatives and Differentiability and the Total Differential

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Let P(x1, x2,…,xn) be a point in Rn, and let f be a function of the n variables x1, x2,…xn. Then the partial derivative of f with respect to xk is that function, denoted by Dkf, such that its function at any point P in the domain of f is given by Dkf(x1,x2,…,xn) = ∆lim x →0 k

f ( x1 , x 2 ,..., x k + ∆x k ,..., x n ) − f ( x1 , x 2 ,..., x n ) ∆x k

if this limit exists. In particular, if f is a function of the three variables x, y and z, then the partial derivatives of f are given by f ( x + ∆x, y , z ) − f ( x, y , z ) D1f(x,y,z) = lim ∆x →0 ∆x f ( x, y + ∆y, z ) − f ( x, y , z ) D2f(x,y,z) = ∆lim y →0 ∆y f ( x, y, z + ∆z ) − f ( x, y , z ) D3f(x,y,z) = lim ∆z →0 ∆z if these limits exist.

 Example 6. Given f(x,y,z) = x2y + yz2 + z3, verify that xf1(x,y,z) + yf2(x,y,z) + zf3(x,y,z) = 3f(x,y,z).

Solution: Holding y and z constant we get

Holding x and z constant we obtain f2(x,y,z) = x2 + z2

f1(x,y,z) = 2xy Holding x and y constant, we get Therefore:

xf1(x,y,z) + yf2(x,y,z) + zf3(x,y,z)

f3(x,y,z) = 2yz + 3z2

= x(2xy) + y(x2 + z2) + z(2yz + 3z2) = 2x2y + x2y +yz2 + 2yz2 + 3z3 = 3(x2y +yz2 + z3) = 3f(x,y,z)

If f is a constant function of two variables, then in general D1f and D2f are also functions of two variables. And if the partial derivatives of these functions exist, they are also called second partial derivatives of f. In contrast, D1f and D2f are called first partial derivatives of f. There are four second partial derivatives of a function of two variables. If f is a function of the two variables x and y, the notations ∂2 f ∂ y∂ x

D2(D1f) D1 2f f1 2 fxy all denote the second partial derivative of f that is obtained by first partial differentiating f with respect to x and then partial-differentiating the result with respect to y. This second partial derivative is defined by f1 ( x, y + ∆y ) − f1 ( x, y ) f1 2(x,y) = ∆lim (7) y →0 ∆y if this limit exists. The notations

∂2 f ∂x 2

D1(D1f) D1 1f f1 1 fxx all denote the second partial derivative of f that is obtained by partial differentiating twice with respect to x. We have the definition f1 1(x,y) = lim

∆x →0

f1 ( x + ∆x, y ) − f1 ( x, y ) ∆x

(8)

if this limit exists. The other two second partial derivatives are defined in an analogous way. f2 1(x,y) = lim

∆x →0

f 2 ( x + ∆x, y ) − f 2 ( x, y ) ∆x

(9)

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f2 2(x,y) = ∆lim y →0

f 2 ( x, y + ∆y ) − f 2 ( x, y ) ∆y

(10)

if these limits exist. The definitions of higher-order partial derivatives are similar. Again there are various notations for a specific derivative. For example, ∂3 f ∂3 f D1 1 2f f1 1 2 fx x y ∂ y∂ x∂ x ∂y∂x 2 all stand for the third partial derivative of f that is obtained by partial-differentiating twice with respect to x and then once with respect to y. In the subscript notation, the order of partial differentiation is from left to right; in the notation

∂3 f , the order is from right ∂ y∂ x∂ x

to left.

 Example

7. Given

f(x,y) = ex sin y + ln xy.

Find (a) D1 1f(x,y);

(b) D1 2f(x,y);

(c)

∂3 f ∂x∂y 2

Solution D1f(x,y) = ex sin y + = ex sin y + (a)D1 1f(x,y) = = ex sin y +

1 ( y) xy

1 x

1 x2

(b)D1 2f(x,y) = ex cos y ∂3 f (c) To find we partial-differentiate twice with respect to y and then once with ∂x∂y 2 respect to x. We have, then, ∂f 1 ∂2 f 1 ∂3 f x = e x cos y + = − e sin y − = −e x sin y ∂y y ∂y 2 y2 ∂x∂y 2 Higher-order partial derivatives of a function of n variables have definitions that are analogous to the definitions of higher-order partial derivatives of two variables. If f is a function of n variables, there may be n2 second partial derivatives of f at a particular point. That is, for a function of three variables, if all the second partial derivatives exist, there are nine of them: f1 1, f1 2, f1 3, f2 1, f2 2, f2 3, f3 1, f3 2, and f3 3.

 Example 8. Find D1 3 2f(x,y,z) if f(x,y,z) = sin (xy + 2z) Solution D1f(x,y,z) = y cos (xy + 2z) D1 3f(x,y,z) = -2y sin (xy + 2z) D1 3 2f(x,y,z) = -2 sin (xy + 2z) – 2xy cos (xy + 2z)

 Example 9. Given f(x,y) = x3y – y cosh xy. Solution

Find (a) fx y(x,y); (b) fy x(x,y)

(a) fx(x,y) = 3x2y – y2 sinh xy fx y(x,y) = 3x2 – 2y sinh xy – xy2 cosh xy (b) fy(x,y) = x3 – cosh xy – xy sinh xy fy x(x,y) = 3x2 – y sinh xy – y sinh xy – xy2 cosh xy = 3x2 – 2y sinh xy – xy2 cosh xy

Observed in Example 9 the “mixed” partial derivatives fx y(x,y) and fy x(x,y) are equal. So for this particular function, when finding the second partial derivative with respect to x and y, the order of differentiation is immaterial. This condition holds for many functions. However, the following example shows that it is not always true.

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 Example 10. Find f1 2(0,0) and f2 1(0,0) if 2 2  (x − y ) ( xy ) 2   x + y2 f ( x, y ) =    0 

if (x,y) ≠ (0,0) if (x,y) = (0,0)

Solution: In example 5, we showed that for this function f1(0,y) = -y for all y and f2(x,0) = x for all x From (7), f1 2(0,0) = ∆lim y →0

f 1 (0,0 + ∆y ) − f 1 (0,0) ∆y

But from (11), f1(0,Δy) = -Δy and f1(0,0) = 0; so − ∆y − 0 f1 2(0,0) = ∆lim y →0 ∆y lim ( − 1) = ∆y →0

= -1 From (9), f2 1(0,0) = lim

∆x →0

f 2 (0 + ∆x,0) − f 2 (0,0) ∆x

From (12), f2(Δx,0) = Δx and f2(0,0) = 0. Therefore ∆x − 0 f2 1(0,0) = lim ∆x →0 ∆x

lim 1 = ∆ x →0 =1 For the function in Example 10 the mixed partial derivatives f1 2(x,y) and f2 1(x,y) are not equal at (0,0). A set of conditions for which f1 2(x0,y0) and f2 1(x0,y0) are equal is given by Theorem 12.3.3, which follows. The function of example 10 does not satisfy the hypothesis of this theorem because both f1 2 and f2 1 are discontinuous at (0,0).

12.3.3 Theorem Suppose that f is a function of two variables x and y defined on an open disk B((x0,y0);r) and fx, fy,fxy and fyx also are defined on B. Furthermore, suppose that fxy and fyx are continuous on B. Then

fxy(x0,y0) = fyx(x0,y0). As a result of this therefore, if the function f of two variables has continuous partial derivatives on some open disk, then the order of partial differentiation can be changed without affecting the result; that is, D1 1 2f = D1 2 1f = D2 1 1f D1 1 2 2f = D1 2 1 2f = D1 2 2 1f = D2 1 1 2f = D2 1 2 1f = D2 2 1 1f and so forth. In particular, assuming that all of the partial derivatives are continuous on some open disk, we can prove that D2 1 1f = D1 1 2f by applying Theorem 12.3.3 repeatedly. Doing this we have D2 1 1f = D1(D2 1f) = D1(D1 2f) = D1[D2(D1f)] = D2[D1(D1f)] = D2(D1 1f = D1 1 2f DIFFERENTIABILITY AND THE TOTAL DIFFERENTIAL We shall define differentiability of functions of more than one variable by means of an equation involving the increment of a function. To motivate this definition we first obtain Partial Derivatives and Differentiability and the Total Differential

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a representation for the increment of a function of a single variable that is similar to what will appear in our definition (12.4.2) of differentiability. Recall from Section 2.1 that f is differentiable function of x and y = f(x), then

∆y ∆x where Δx and Δy are increment of x and y and Δy = f(x + Δx) – f(x) When │Δx│is small and Δx ≠ 0, Δy/Δx differs from f ′(x) by a small number that depends on Δx, which we shall denote by є. Thus ∆y − f ' ( x) є= if Δx ≠ 0 ∆x where є is a function of Δx. From this equation we obtain Δy = f ′(x) Δx + є Δx where є is a function of Δx and є→0 as Δx→0 From the above it follows that if the function f is differentiable at x0, the increment of f at x0, denoted by Δf(x0), is given by f ′(x) = lim

∆x →0

lim є = 0 Δf(x0) = f ′(x0) Δx + є Δx where ∆ x →0 For the functions of two or more variables an equation corresponding to this one is used to define differentiability. And from the definition we determine criteria for a function to be differentiable at a point. We give the details for a function of two variables and begin by defining the increment of such a function.

Definition of an Increment of a Function of Two Variables If f is a point of two variables x and y, then the increment of f at the point (x0,y0), denoted by Δf(x0,y0), is given by Δf(x0,y0) = f(x0 + Δx, y0 + Δy – f(x0,y0)



Illustration 1. For the function defined by f(x,y) = 3x – xy2

We find the increment of f at any point (x0,y0)

Δf(x0,y0) = f(x0 + Δx, y0 + Δy – f(x0,y0) = 3(x0 + Δx) – (x0 + Δx)( y0 + Δy)2 – (3x0 – x0y02) = 3x 0 + 3 Δx – x0y02 – y02 Δx – 2x0y0 Δy – 2 y0 Δx Δy – x0(Δy)2 – Δx(Δy)2 –

3x0 + x0y02

= 3 Δx – y02 Δx – 2x0y0 Δy – 2 y0 Δx Δy – x0(Δy)2 – Δx (Δy)2

12.4.2 Definition of an Increment of a Function of Two Variables

If f is a function of two variables x and y and the increment of f at (x0,y0) can be written as Δf(x0,y0) = D1f(x0,y0) Δx + D2f(x0,y0) Δy + є1 Δx + є2 Δy where є1 and є2 are functions of Δx and Δy such that є1 →0 and є2 →0 as (Δx, Δy)→(0,0), then f is differentiable at (x0,y0).

 Illustration 2 We use Definition 12.4.2 to prove that the function of 1 is differentiable at all points in R2. We must show that for all points (x0,y0) in R2 we can find an є1 and an є2 such that Δf(x0,y0) = D1f(x0,y0) Δx + D2f(x0,y0) Δy + є1 Δx + є2 Δy and є1 →0 and є2 →0 as (Δx, Δy)→(0,0). Because f(x,y) = 3x – xy2, D1f(x0,y0) = 3 – y02 and D2f(x0,y0) = -2 x0y0 With these values and the value of Δf(x0,y0) from illustration 1,

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Δf(x0,y0) – D1f(x0,y0) Δx – D2f(x0,y0) Δy = –x0(Δy)2 – 2 y0 Δx Δy Δx (Δy)2 The right side of the above equation can be written in the following ways: [ -2y0 Δy - (Δy)2 ] Δx + (-x0 Δy) Δy ( -2y0 Δy ) Δx + ( - Δx Δy – x0 Δy ) Δy [ -( Δy)2 ] Δx + (-2y0 Δx – x0 Δy ) Δy 0 ∙ Δx + [ -2y0 Δx – Δx Δy – x0 Δy ] So there are at least four possible pairs є1 = –2y0Δy – (Δy)2 and є1 = –2y0 Δy and є1 = – (Δy)2 and є1 = 0 and

of values for є1 and є2: є2 = –x0 Δy є2 = – Δx Δy –x0 Δy є2 = –2y0 Δx –x0 Δy є2 = –2y0 Δx – Δx Δy – x0 Δy

For each pair,

lim

( ∆x , ∆y )→( 0 , 0 )

lim

є1 = 0

( ∆x , ∆y )→( 0 , 0 )

є2 = 0

Note that it is only necessary to find one pair of values for є1 and є2. The next theorem states that for a function of two variables, differentiability implies continuity, just as was the case for a function of single variable.

12.4.3 Theorem

If a function of two variables is differentiable at a point, it is continuous at that

point. For a function f of a single variable, the existence of the derivative of f at a number implies differentiability and, therefore, continuity at that number. However, for a function of two variables the existence of the partial derivatives at a point does not imply differentiability at that point.



 xy  x2 + y2 f ( x, y ) =     0

Example 1. Given

if (x,y) ≠ (0,0) if (x,y) = (0,0)

prove that D1f(0,0) and D2f(0,0) exist but that f is not differentiable at (0,0). Solution: f (0, y ) − f (0,0) f ( x,0) − f (0,0) D1f(0,0) = lim D2f(0,0) = lim y →0 x →0 y −0 x −0 0 −0 0 −0 = lim = lim y → 0 x →0 y x = lim 0 = lim 0 y →0

x →0

=0 Therefore both D1f(0,0) and D2f(0,0) exist. In

lim

( x , y ) →( 0 , 0 )

Example

6 of

Section

12.2

=0 we

demonstrated

that for this function

f ( x, y ) does not exist; hence f is not continuous at (0,0). Because f is not

continuous at (0,0), then Theorem 12.4.3 is not differentiable there. While the mere existence of the partial derivatives of a function of two variables at a point does not guarantee differentiability at that point, there are additional conditions required of the function that provide such a guarantee. These conditions are given in the following theorem, whose proof appears in the supplement of this section.

12.4.4 Theorem

Let f be a function of x and y such that D1f and D2f exist on an open disk B(P0;r), where P0 is the point (x0,y0). Then if D1f and D2f are continuous at P0, f is differentiable at P0. Partial Derivatives and Differentiability and the Total Differential

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This theorem is much easier to apply than Definition 12.4.2 to prove differentiability of a function of two variables. For instance, because the partial derivatives of any polynomial function are also polynomials and polynomial functions are continuous everywhere, Theorem 12.4.4 tells us that polynomial functions are differentiable everywhere.



Example 2 Use Theorem 12.4.4 to prove that the function defined by f(x,y) = xey – y ln x is differentiable at all points in its domain.

Solution The domain of f is the set of all points (x,y) in R2 for which x > 0. We compute the partial derivatives: y D1f(x,y) = ey – D2f(x,y) = xey – ln x x Because D1f and D2f are continuous at all points in R2 for which x > 0, then from Theorem 12.4.4, f is differentiable at all points in its domain. A function satisfying the hypothesis of Theorem 12.4.4 at a point is said to be continuously differentiable at the point. Whereas continuous differentiability at a point is a sufficient condition to prove that a function is differentiable at a point, it is not a necessary one. That is, it is possible for a function to be differentiable at a point even if its partial derivatives are not continuous there The equation in Definition 12.4.2 is Δf(x0,y0) = D1f(x0,y0) + D2f(x0,y0) Δy + є1 Δx + є2 Δy

(2)

The expression involving the first two terms on the right side of this equation is called the principal part of Δf(x0,y0) or the total differential of f at (x0,y0).

12.4.5 Definition of the Total Differential of a Function of Two Variables If f is a function of two variables x and y, and f is differentiable at (x,y), then the total differential of f is the function df having function values given by df(x,y, Δx, Δy) = D1f(x,y) Δx + D2f(x,y) Δy Note that df is a function of the four variables x, y, Δx and Δy. If z = f(x,y), we sometimes use dz in place of df(x,y, Δx, Δy), and write dz = D1f(x,y) Δx + D2f(x,y) Δy

(3)

If in (3) , f(x,y) = x, then z = x, D1f(x,y) = 1, and D2f(x,y) = 0; so (3) gives dz = Δx. Because z = x, for this function dx = Δx. In a similar fashion, if we take f(x,y) = y, then z = y, D1f(x,y) = 0, and D2f(x,y) = 1; thus (3) gives dz = Δy. Because z = y, then for this function dy = Δy. Hence we define the differentials of the independent variables as dx = Δx and dy = Δy. Then (3) can be written as dz = D1f(x,y) dx + D2f(x,y) dy (4) and at the point (x0,y0), dz = D1f(x0,y0) dx + D2f(x0,y0) dy (5) In (2) let Δz = Δf(x0,y0), dx = Δx and dy = Δy. Then Δz = D1f(x0,y0) dx + D2f(x0,y0) dy + є1 dx + є2 dy By comparing this equation and (5), observe that when dx (i.e., Δx) and dy (i.e., Δy) are close to zero, and because then є1 and є2 also will be close to zero, dz is an approximation to Δz. ∂ z ∂z Equation (4) with the notation and becomes ∂ y ∂x

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dz =

∂z ∂z dx + dy ∂x ∂y

(6)

 Example 1 A closed metal container in the shape of a right-circular cylinder has an inside height of 6 in., an onside radius of 2 in., and a thickness of 0.1 in. if the cost of the metal is 40 cents per cubic inch, approximate by differentials the total cost of the metal used in the manufacture of the container. Solution If V cubic inches is the volume of the right-circular cylinder of radius r inches and height h inches, then V = πr 2 h The exact volume of metal in the container is the difference between the volumes of two right-circular cylinders for which r = 2.1, h = 6.2 and r = 2, h = 6, respectively. The increment ΔV gives the exact volume of metal, but because only an value is wanted, we find dV instead. From (6) ∂V ∂V dr + dh ∂r ∂h = 2rh dr + r2 dh

dV =

With r = 2, h = 6, dr = 0.1, and dh = 0.2 dV = 2(2)(6)(0.1) + (2)2 (0.2) = 3.2 Thus ΔV ≈ 3.2, so that the volume of metal in the container is approximately 3.2 in3. Because the cost of the metal is 40 cents per cubic inch, the number of cents in the approximate cost of the metal is, therefore,128 ≈ 402. Conclusion: The approximate cost of the metal is $4.02.

12.4.6 Definition of an Increment of a Function of n Variables _

_

_

_

If f is a function of the n variables x1, x2,…xn, and P is the point ( x 1 , x 2 ,... x n ) , _

then the increment of f at P is given by _

_

_

_

_

∆f ( P ) = f ( x1 + ∆x1 , x 2 + ∆x 2 ,..., x n + ∆x n ) − f ( P )

12.4.7 Definition of a Differentiable Function of n Variables If f is a function of the n variables x1, x2,…xn, and the increment of f at the point _

P can be written as _

_

_

_

Δf( P ) = D1 f( P ) Δx1 + D2 f( P ) Δx2 + Dn f( P ) Δxn + є1 Δx1 + є2 Δx2 + … + єn Δxn where є1 →0, є2 →0,…, єn →0, as (Δx1, Δx2,…Δxn) → (0,0,…0), then f is said to be _

differentiable at P .

12.4.8 Definition of the Total Differential of a Function of n Variables

If f is a function of the n variables x1, x2,…xn, and differentiable at P, then the total differential of f is the function df having function values given by Partial Derivatives and Differentiability and the Total Differential

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_

df(P, Δx1, Δx2,…Δxn) = D1 f(P) Δx1 + D2 f(P) Δx2 + . . . + Dn f( P ) Δxn Letting w = f(x1, x2,…xn), defining dx1 = Δx1, dx2 = Δx2, … , dxn = Δxn, and using the ∂w dx 1 instead of Di f(P), we can write the equation of Definition 12.4.8 as notation ∂x1 ∂w ∂w ∂w dx 1 + dx 2 + . . . dx n dw = (7) ∂x1 ∂x 2 ∂x n

 Example 4 Measurements of the dimensions of a box are 10cm, 12cm and 15cm, each correct to 0.02cm. (a) Approximate by differentials the greatest error if the volume of the box is computed from these measurements. (b) Approximate the percent error Solution

(a)If V cubic centimeters is the volume of the box whose dimensions are x, y, and z centimeters, V = xyz

The exact value of the error is found approximation to ΔV. From (7) for three independent ∂V ∂V dy + dx + dV = ∂y ∂x

from ΔV; however, we use dV as an variables, ∂V dz ∂z

= yz dx + xz dy + xy dz

From the given facts │Δx│≤ 0.02, │Δy│≤ 0.02, and │Δz│≤ 0.02. To find the greatest error in the volume we take the greatest error in the measurements of the three dimensions. So taking dx = 0.02, dy = 0.02, dz = 0.02, and x=10, y = 12, z = 15, we have dV = (12)(15)(0.012) + (10)(15)(0.02) + (10)(12)(0.02) =9 Thus ΔV ≈ 9. Conclusion: The greatest possible error in the calculation of the volume from the given measurements is approximately 9 cm3. (b)The relative error is found by dividing the error by the actual value. Therefore, the relative error in computing the volume from the given measurements is ΔV/V ≈ dV/V. Because dV/V = 9/1800 ∆V ≈ 0.005 V Conclusion: The approximate percent error is 0.05 percent.

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